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spayn [35]
4 years ago
9

A 42 N net force is applied to a box which slides horizontally across a floor for 9.0

Physics
1 answer:
Rufina [12.5K]4 years ago
8 0

Answer:

The Amount of work is done on the box by the net force is 378 Joules.

Explanation:

Given:

A 42 N net force is applied to a box which slides horizontally across a floor.

Force = 42 N

Force applied for a displacement of 9.0 m

Displacement = 9.0 m

To Find:

Amount of work is done on the box by the net force = ?

i.e Work = ?

Solution:

When a force causes a object to move, work is being done on the object by the force. Work is the measure of energy transfer when a force (F) moves an object through a distance (d).

Formula is given by

Work=Force\times Displacement

Substituting the given values we get

∴ Work = 42 × 9.0

∴ Work = 378 N-m

or

∴ Work = 378 Joules

The Amount of work is done on the box by the net force is 378 Joules.

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A lady walks 10 m to the north, then she turns and continues walking 30 m due east.
ANEK [815]

Answer:

The distance covered is 40 m and the displacement is 31,6m.

Explanation:

The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .

4 0
3 years ago
How do you calculate the braking distance
Anettt [7]

The braking distance is given by s=\frac{-u^2}{2a}

Explanation:

When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

v^2-u^2=2as

where

u is the initial speed of the car

v is the final speed of the car, which is zero because the car comes to rest:

v = 0

a is the acceleration of the car

s is the distance travelled by the car during the deceleration, so it is the braking distance

Therefore, re-arranging the equation for s, we find an expression for the braking distance:

s=\frac{-u^2}{2a}

Note that the sign of a is negative since the car is decelerating, therefore the final sign of s is positive.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2562700

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6 0
4 years ago
If the same change in velocity occurs in less time, does the magnitude of the corresponding average acceleration increase, decre
Zielflug [23.3K]

Answer:

increase

Explanation:

As acceleration is the time rate of change of velocity

a = Δv/t

with t in the denominator, smaller time values result in greater acceleration values with constant velocity change.

7 0
3 years ago
the young’s modulus of aluminum is 69gpa, of nylon is 3gpa, of tungsten is 400gpa, and of copper is 117gpa. if equal-size sample
olga_2 [115]

Answer:

Least to most elongated: tungsten, copper, aluminum, nylon.

Explanation:

Materials with high Young's modulus are difficult to stretch. σ = Yε and ε = ΔL/L so an object with a high Young's modulus (Y) subject to a certain tensile stress (σ) will have a smaller strain than an object with a smaller Young's 's modulus subject to the same tensile stress. If strain (ε) is smaller, then ΔL will also be smaller.

3 0
2 years ago
A car with mass mc = 1490 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck o
icang [17]

Answer:

v= - 4.507 i - 2.363 j

Explanation:

 Given that

mc= 1490 kg

vc= 9.5 m/s ( - i)

mt=  1650 kg

vt = 6.4 m/s ( -j)

There is any external force so linear momentum will remain conserve.

Lets take final speed is v.

mc .vc + mt . vt = ( mc+mt) v

1490 x 9.5 ( - i) + 1650 x 6.4 ( -j) = ( 1490+1650) v

14,155 ( -i) + 10,560 ( - j) = 3140 v

v= - 4.507 i - 2.363 j

3 0
4 years ago
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